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Math Help - ineqaulity

  1. #1
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    ineqaulity

    Can anybody help me with the proof of this inequality please?
     <br />
\sum_{k=1}^n(a_k) \cdot \sum_{k=1}^n(1/a_k) \leq n^2<br />

    Thank you
    Last edited by mr fantastic; September 30th 2009 at 03:33 AM. Reason: Improved latex
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  2. #2
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    There are probably other ways to approach this question, but here's my attempt at a proof by induction. I'll skip the S(1) and S(k) steps for convenience and go on to S(k+1):

    (a_1+a_2+...+a_k+a_{k+1})(\frac{1}{a_1}+\frac{1}{a _2}+...\frac{1}{a_k}+\frac{1}{a_{k+1}})\\=(a_1+a_2 +...+a_k)(\frac{1}{a_1}+\frac{1}{a_2}+...\frac{1}{ a_k})+a_{k+1}(\frac{1}{a_1}+\frac{1}{a_2}+...\frac {1}{a_k})+\frac{1}{a_{k+1}}(a_1+a_2+...+a_k)\\\leq n^2+(\frac{a_1}{a_{k+1}}+\frac{a_{k+1}}{a_1})+(\fr ac{a_2}{a_{k+1}}+\frac{a_{k+1}}{a_2})+...(\frac{a_ k}{a_{k+1}}+\frac{a_{k+1}}{a_k})+\frac{a_{k+1}}{a_ {k+1}}\\\leq k^2+2k+1\,\,$since$\,(\sqrt{\frac{x}{y}}-\sqrt{\frac{y}{x}})^2\geq 0\Rightarrow \frac{x}{y}+\frac{y}{x}\geq 2\\\leq (k+1)^2

    Sorry, latex thing doesnt seem to be working
    Last edited by vuze88; October 3rd 2009 at 02:51 AM.
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  3. #3
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    a_1=1, a_2=2 \Rightarrow \sum_{k=1}^{2}a_k \cdot \sum_{k=1}^{2}\frac{1}{a_k} =(1+2)\cdot(\frac{1}{1} + \frac{1}{2}) = 3\cdot 1.5 = 4.5 > 4 = n^2
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  4. #4
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    Quote Originally Posted by sidi View Post
    Can anybody help me with the proof of this inequality please?
     <br />
\sum_{k=1}^n(a_k) \cdot \sum_{k=1}^n(1/a_k) \leq n^2<br />

    Thank you
    I believe you have the inequality reversed in the problem statement.

    I think you will also need the additional hypothesis that a_k > 0 for all k.
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